CHAPTER 1A 1.1(a) Could 13 g of xenon gas in a vessel of volume 10 dm3 exert a pressure of 20 atm at 25C if it behaved as a gas ? If not, what pressure would it exert? (b) What pressure would it exert if it behaved as a Vander Waals gas? Given: 13 g Xe of 131.29g/mol MW V = 1 dm3 = 1L P = 20atm T = 25C Solution: a. For the number of moles: n= mass / molar mass n = 13 131.29 n = 0.0990 mol For the compressibility test: Z = PV nRT Z = (20atm)(1L) (0.0990mol)(0.08206L atm/mol K)(298..15K) Z = 8.2571

Therefore, it is not an ideal gas. Then, P = nRT V P = (0.0990mol)(0.08206L atm/mol K)(298.15K) 1L P = 24.4662 atm b. P = (RT/ V-b) - (a/V2) P = [(0.08206Latm/mol K)(298.15K)/ (1L 5.16x10-2 dm3 /mol)] - (4.137atm dm6 /mol 2/1L) P = 22 atm

1.2(a) A perfect gas undergoes isothermal compression, which reduces its volume by 2.20 dm3. The final pressure and volume of a gas are 5.04 bar and 4.65 dm3, respectively. Calculate the original pressure of the gas in (a) bar, (b) atm. Given: V = 2.20 dm3 P2 = 5.04 bar V2 = 4.65 dm

3 Solution: a. P1V1 = P2 V 2 P1 = (4.65 dm

3)( 5.04 bar) (2.20 + 4.65)dm3 P1 = 3.4213 atm b. P1 = 3.4213atm x (1bar/1.01325atm) P1 = 3.3766 bar

1.3(a) A car tyre (i.e. an automobile tire) was inflated to a pressure of 24 lb in -2 (1.00 atm=14.7 lb in -2) on a winters day when the temperature was -5C. What pressure will be found, assuming no leaks have occurred and that the volume is constant, on a subsequent summers day when the temperature is 35C? What complications should be taken into account in practice? Given: P1 = 24psi T1 = -5C T2 = 35 Solution: P1/ T1 = P2/ T2 P2 = (24psi x 308.15K) 268.15K P2 = 27.5801 psi 1.4(a) A sample of 255 mg of neon occupies 3.00 dm3 at 122 K. Use the perfect gas law to calculate the pressure of the gas. Given: Mass = 255 mg V = 3.00 dm3

T = 122K Solution: P = nRT V P = [(0.255g/20.179g/mol)(0.08206 Latm/molK)(122K)] 3.00 dm3 P = 0.0422 atm 1.8 (a). At 500C and 93.2kPa, the mass density of a sulphur vapour is 3.710kg/m3. What is the molecular formula of the sulphur vapour under this condition? Given: T = 500C P = 93.2kPa D = 3.710kg/m3

Solution: D = PM RT M = (8.314Pa m3 /mol)(773.15IK)( 3.710kg/m3) 9.32x104 Pa M = 0.2565 kg/mol M = 256.5 g/mol The number of atoms per molecule is: # = 256.6g/mol 32g/mol # = 8

The formula is S8.

1.10(a) Given that the density of air at 0.987 bar and 27C is 1.146 kg m-3 , calculate the mole fraction and partial pressure of nitrogen and oxygen assuming that (a) air consist of only of these two gases, (b) air also contains 1.0 mole per cent Ar. Given: P = 0.987 bar = 0.9741 atm T = 27C D = 1.146 kg/m3

Solution: PV = nRT V = [(0.08206 Latm/molK)(303.15K)] 0.9741atm V = 25.5379 L D = m/V m = 1.146 kg/m3 x 25.5379 L m = 28.98 g a. Let x = N2 1 - x = O2 x (28.02) + (1-x)(32) = 28.98 -3.98x = 28.98 32 x = 0.759 1 x = 0.241 PN2 = 0.759 x 0.987 bar PN2 = 0.7491 bar PO2 = 0.241 x 0.987 bar PN2 = 0.2379 bar b. Let Ar = 0.01 x = N2 0.99 - x = O2 x (28.02) + (0.99 - x)(32) + 0.01(39.95) = 28.98 -3.98x = 28.98 31.68 0.3995 x = 0.7798 0.99 x = 0.2102 PN2 = 0.7798 x 0.987 bar PN2 = 0.770 bar PO2 = 0.2102 x 0.987 bar PN2 = 0.2075 bar PAr = 0.01 x 0.987 bar PAr = 0.00987 bar

1.11(a) The density of a gaseous compound was found to be 1.23 kg m-3 at 330 K and 20 kPa. What is the molar mass of a compound? Given: D = 1.23 kg m-3 T = 330K P = 20kPa = 0.1974 atm Solution: D = PM RT M = [(0.08206 Latm/molK)(330K)( 1.23 kg m-3) 0.1974atm M = 168.7343 g/mol 1.13(a) Calculate the pressure exerted by 1.0 mol C2H6 behaving as (a) a perfect gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 273.15 K in 22.414 dm3, (ii) at 1000 K in 100 cm3. Use the data in table 1.6. Given: n = 1.0 mol C2H6 i. 273.15 K in 22.414 dm3 ii. 1000 K in 100 cm3

Solution: PV = nRT a. P = [(1mol)(0.08206 Latm/molK)(273.15K)] 22.414 dm3 P = 1.000atm P = [(1mol)(0.08206 Latm/molK)(1000K)] 0.1 dm3 P = 820.6 atm b. (P + (n2a/V2 )(V - nb) = nRT direct substitution to the formula,we have P = 0.991atm P = 1800.5894 atm 1.14(a) Express the van der Waals parameters a=0.751 atm dm6 /mol2 and b = 0.0226 dm3/mol in SI base units. Solution: a. 0.751 atm dm6 /mol2 = 0.07 kg m5/ mol2 s 2

b. 0.0226 dm3/mol = 2.26 x 10-5 m3 /mol 1.15(a) A gas at 250 K and 15 atm has a molar volume 12 per cent smaller than that calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces?

Given: T = 250 K P = 15 atm Solution: PV = RT V = [(0.08206 L atm/mol K)(250K)] 15 atm V = 1.3677 L/mol V = [1.3677 x (.12 x 1.3677)] L/mol V = 1.2036 L/mol

The dominating one is the attractive forces. Z = [(15atm)(1.2036 L/atm)] [(0.08206 L atm/mol K)(250K)] Z = 0.8800 1.17(a) Suppose that 10.0 mol C2H6 (g) is confined to 4.860 dm' at 27C. Predict the pressure exerted by the ethane from (a) the perfect gas and (b) the van der Waals equations of state. Calculate the compression factor based on these calculations. For ethane, a = 5.507 atm dm6 /mol2, b=0.0651 dm3/mol. Given: n = 10mol V = 4.860 dm' T = 27C. Solution: a. PV = nRT P = [(10mol)(0.08206 L atm/mol K)(300.15K)] 4.860 dm' P = 50.6796 atm b. P + (n2a/V2 )(V - nb) = nRT direct substitution to the formula, we have P = 35.2028 atm Z = PV nRT Z = [(35.2028 atm)( 4.860 dm3)] [(10mol)(0.08206 L atm/mol K)(300.15K)] Z = 0.6946 1.18a) A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K. Calculate (a) the mole fractions of each component, (b) their partial pressures, and (c) their total pressure. Given: V = 22.4 dm3 n = 2.0 mol H2 and 1.0 mol N2 T = 273.15K

Solution: Mole fraction: a. H2 = 2.0/3.0 = 0.6667 N2 = 1.0/3.0 = 0.3333 PV = nRT P = [(3mol) (0.08206 L atm/mol K) (273.15K)] P = 3.0020 atm b. H2 = 0.6667 x 3.0020 atm = 2.0014 atm N2 = 0.3333 x 3.0020 atm = 1.0006 atm c. P = 2.0014 + 1.0006 = 3.002 atm 1.19(a) The critical constants of methane are Pc= 45.6 atm, Vc = 98.7 cm3/mol and Tc = 190.6 K. Calculate the van der Waals parameters of the gas and estimate the radius of the molecules. Given: Pc= 45.6 atm Vc = 98.7 cm3/mol Tc = 190.6 K Solution: a. Vc = 3b b = 98.7 cm3/mol / 3 b = 32.9 cm3/mol

b. r = 0.5 (3b/4N)1/3

r = 0.5[(3)(32.9 cm3/mol) 4 x 6.022x1023 /mol]1/3 r = 1.77 x 10-8 cm c. a = 27Pcb2 a = 27 x 45.6 atm x (32.9 cm3/mol) 2

a = 1.3327 atm dm6 /mol2 1.20(a) Use the van der Waals parameters for chlorine to calculate approximate values of (a) the Boyle temperature of chlorine and (b) the radius of a CI 2 molecule regarded as a sphere. Given: a = 6.260 atm dm6 /mol2 b = 5.42 x 10-2 dm3/mol Solution: T = a / Rb T = 6.260 atmdm6 /mol2 (0.08206 L atm/mol K)( 5.42 x 10-2 dm3/mol ) T = 1407.4842K

b. r = 0.5 (3b/4N)1/3

r = 0.5[(3)( 5.42 x 10-2 dm3/mol) 4 x 6.022x1023 /mol]1/3

r = 1.39 x 10-9 dm 1.21(a) Suggest the pressure and temperature at which 1.0 mol of (a) NH" (b) Xe, (c) He will be in states that correspond to 1.0 mol Hz at 1.0 atm and 25e. Given: H2 NH3 Xe He Pc 12.8 atm 111.3 atm 58 atm 2.26 atm Tc 33.23 K 405.5 K 289.75 K 5.21 K Solution: For H2 ] Pr = 1 atm/ 12.8 atm Tr = 298.15K / 33.23K Pr = 0.0781 Tr = 8.9723 a. NH3 P = Pr x Pc T = Tr x Tc P = 0.0781 x 111.3 atm T = 8.9723 x 405.5 K P = 8.6925 atm T = 3638.2677 K b. Xe P = Pr x Pc T = Tr x Tc P = 0.0781 x 58 atm T = 8.9723 x 289.75 K P = 4.5298 atm T = 2599.7239 K b. He P = Pr x Pc T = Tr x Tc P = 0.0781 x 2.26atm T = 8.9723 x 5.21K P = 0.1765 atm T = 46.7457 K 1.22(a) A certain gas obeys the van der Waals equation with a = 0.50 Pa m6 /mol2. Its volume is found to be 5.00 X 10-4 m3/mol at 273 K and 3.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? Given: T = 273.0 K P = 3MPa a = 0.50 Pa m6 /mol2 V = 5.00 X 10-4 m3/mol Solution: a. P + (n2a/V2) (V - nb) = nRT direct substitution to the formula, we have b = 4.6056 x 10-5 m3/mol b. Z = PV / RT Z = (3MPa)( 5.00 X 10-4 m3/mol) (8.314 Pa/mol K)(273 K) Z = 0.6609

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